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Haryana PMT Haryana PMT Solved Paper-2001

  • question_answer
    In a compound \[C=40%\]and \[H=6.66%,\] what is the empirical formula of the compound?

    A)                

    B)  \[{{C}_{4}}H{{O}_{4}}\]

    C)  CHO                     

    D)  \[C{{H}_{2}}O\]

    Correct Answer: D

    Solution :

                    Percentage \[O%=100-(40+6.66)\]\[=53.34%\]
                    Symbol % Atomic mass   Relative  no. of atoms Simple Reaction
    C 40 12 \[\frac{40}{12}=3.33\] \[\frac{3.33}{3.33}=1\]
    H 6.66 1 \[\frac{6.66}{1}=6.66\] \[\frac{6.66}{3.33}=2\]
    O 53.34 16 \[\frac{53.34}{16}3.33\] \[\frac{3.33}{3.33}=1\]
                 Empirical formula \[=C{{H}_{2}}O\]


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